Wiskunde Multiple Choice Vraag

$\displaystyle\lim_{x\to\,0}\;\frac{\sin^3x}{\sin x-\tan x}=$	A. 1
B. 2
C. 3
D. −3
E. −2

1^ste manier : zonder de regel van De L'Hospital
$\inline\\\displaystyle\lim_{x\to\,0}\;\frac{\sin^3x}{\sin x-\tan x}=\lim_{x\to\,0}\;\frac{\sin^2x}{1-\frac{1}{\cos x}}=\lim_{x\to\,0}\;\frac{\sin^2x.\cos x}{\cos x-1}\\\lim_{x\to\,0}\;\frac{(1-\cos^2x).\cos x}{\cos x-1})=\lim_{x\to\,0}\;\frac{(1-\cos x)(1+\cos x).\cos x}{\cos x-1}\\=-\lim_{x\to\,0}\;(1+\cos x).\cos x=-(1+1).1=-2$
2^de manier : met de regel van De L'Hospital
$\inline\\\displaystyle\lim_{x\to\,0}\;\frac{\sin^3x}{\sin x-\tan x}\left(=\frac00\right)\overset{H}{=}\lim_{x\to\,0}\;\frac{3\sin^2x.\cos x}{\cos x-\frac{1}{\cos^2x}}=3\lim_{x\to\,0}\;\frac{\sin^2x.\cos^3x}{\cos^3x-1}\\=3\lim_{x\to\,0}\;\frac{(1-\cos^2x).\cos^3 x}{\cos^3 x-1})=3\lim_{x\to\,0}\;\frac{(1-\cos x)(1+\cos x).\cos^3 x}{(\cos x-1)(\cos^2x+\cos x+1)}\\=-3\lim_{x\to\,0}\;\frac{(1+\cos x).\cos^3 x}{\cos^2x+\cos x+1}=-3\cdot\frac{(1+1).1^3}{1^2+1+1}=-3\cdot\frac{2.1}{3}=-2$

^{gricha - vA154 - 28.3.2026}